# How can students prognosticate to rewrite the same sum backwards, then add the same sum twice?

This comment doesn't fulfill my students or me, because it doesn't demystify this trick of writing $$S_n$$ forward, then backwards, then adding. What would spur students to action these unnatural steps? How would students vaticinate this trick? I'm hankering after an answer other than "You just have to memorize this trick after you see it"!

Our textbook below also doesn't expatiate this trick!

Problem 3.6: Find a formula for the sum of the first n positive integers.

David Patrick, Introduction to Counting & Probability (2005), p 54.

• Gauss did not have a smartphone to watch 24/7, so he was playing with numbers. Finding patterns and making them into theorems and formulas is what mathematicians do. Jan 25, 2022 at 0:28
• @RustyCore: +1 to your comment. I think I better put my smartphone aside and go to bed now... Jan 25, 2022 at 1:56
– JRN
Jan 25, 2022 at 3:08
• I guess that if we take 1000 people who never saw the sum of an arithmetic progression or any similar proof, and ask them to calculate $1+2+\ldots+100$, little to none of them would come up with the procedure that Gauss purportedly discovered when he was very young. I'm all for giving motivation for proofs when possible, but sometimes the most honest thing to say is that in math we are trying to learn from the cumulative effort of tens of thousands of geniuses. (Just by the sheer amount of cumulative research years many tricks are expected to be discovered, even if it wasn't for genius).
– Snaw
Jan 25, 2022 at 15:25
• Research is unpredictable (or else it would be technique). It is hard to speculate how Gauss discovered this (or how others before him did). Maybe he calculated some sum and then to check himself (remember, no computers) decided to calculate it backwards, and then another time he was bored and decided to calculate it in pairs and noticed the pattern. Maybe the fact that $1+100=2+99$ just happened to pop into his mind. Maybe he was such a genius that the whole thing was obvious to him upon first glance. Maybe he just assumed there had to be an easy way and started playing around with the sum.
– Snaw
Jan 25, 2022 at 15:32

This answer has already more or less appeared in some comments, but it's not expected that every student should come up with this "trick" on their own. I know I didn't. The important thing is that they see that there's nothing especially tricky about it, and given the right circumstances it's something they could have come up with. Most people have noticed the fact that $$1+9=2+8=3+7=\cdots=9+1$$, and from there it's not much of a leap to Gauss' method. Nobody should feel bad if they don't notice a seemingly tricky method right away. What's important is that they understand it (maybe they even think it's beautiful and inspires them to learn math!) and possibly use a similar method later. It doesn't have to be a case of memorizing some incomprehensible rule.

• These kinds of tricks have a similar feeling to a joke to me. You don't have to be able to come up with a joke to "get it" and find it funny enough to want to share. Jan 25, 2022 at 19:02

I think the main message that you need to communicate to your students, since it is not coming from the book, is that genuine problem solving involves several stages. To quote Polya, these are "understanding the problem" (i.e., considering the sum from $$1$$ to $$n$$), "devising a plan" (forming a conjecture of what we believe the sum to equal) , "carrying out the plan" (writing a formal proof of the conjecture), and then "looking back" (understanding our solution and alternatives to take something away for future problems). Even though this is a simple problem, it can still be used to illustrate the stages and cultivate a more realistic and productive attitude toward problem-solving than the initial surprised reaction of "how did I not think of that trick?".

The explanation in the book is a concise path from the problem statement to the conclusion; it does not illustrate or offer much guidance on forming a conjecture, which is a big part of the battle on this problem for a new student (if not so difficult for Gauss). I think with a conjecture in mind, it would be easier to come up with the "reverse the sum" justification.

The natural next question: how does one form a conjecture? Just as commenter Rusty Core said -- by playing with the numbers. It's a little difficult for me to put myself into the shoes of a student who has never seen this identity before, but one approach that came to mind is to mock up some data in a spreadsheet (or calculate a few values by hand) and speculate on a pattern. With the spreadsheet, it should be easy to check that your guess is at least consistent with a large number of cases. From there, you could use a more structured approach like induction to actually prove your conjectured formula.

Another way to form a conjecture is to draw a picture for inspiration... here is a doodle I drew up over coffee:

So I drew a few arrays of shaded circles and then conjectured that it seems you can complete the arrays to be a square (empty circles).

Translating that into symbols: $$\sum_{i=1}^{n} i = n^2 - \sum_{i=1}^{n-1} i$$ which can be seen to be equivalent to the familiar identity, and from there you can use some more structured approach like induction for the formal proof.

Alternatively, you could start by just considering the sum in isolation and then manipulate it into equivalent algebraic forms in the hopes of finding a pattern. This is the solution outlined in the book. Having seen this solution, the reader has the opportunity to "look back" and ask what they can take away from this to future problems? One thing they may not have considered is to not only think of the sum as having components but also to think of it as a whole (i.e., assign it a label, $$S$$). Then you might ask: is there a way to produce a simple formula in terms of $$n$$ performing "simple" operations on $$S$$? If we can limit the operations on $$S$$ to operations with constants or functions of $$n$$, then we may be able to solve resulting equations for $$S$$ in terms of $$n$$. I'm not sure how much total mileage you can get out of this line of thinking in discrete math, but it comes up at least one more time when deriving the formula for the sum of a finite geometric series.

• xkcd: xkcd.com/2569 Jan 27, 2022 at 2:04

Adding to the other good answers, a very-slight variant of that trick is (to me) more understandable and more scale-able to other problems, "once the lesson is learned". Namely, grouping of terms in a large sum. In the example at hand, it's to group first-and-last, second-and-next-to-last, etc. Yes, when the total number of summands is even this pairs everything, and then it's odd the middle term is left over. With an even number $$n$$, we get $${n\over 2}\cdot (n+1)$$. With an odd number $$n$$ we get $${n-1\over 2}\cdot (n+1)+{n\over 2}$$. As we know, these give the same formula...

The grouping idea does also apply to "telescoping" sums, for example.

So, as in other answers, it's both learning re-usable ideas from seeing such things, as well as acquiring methodological approaches to come up with then on one's own.

Just draw it. You will stumble on this method of proof immediately.

Let's say you don't know the formula yet, so you draw $$\sum_{i=1}^5 i$$:

You notice that this shape is a triangle, so maybe it is half of a square like in geometry.

But when you draw another copy of the triangle:

you notice that it didn't form a square after all; in fact it was half of a 5 x 6 rectangle. You conjecture immediately that $$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$ and start to look for an algebra trick to prove it.

But the "trick" is just to look at the rows.

The rows show that you should pair the red 5 with the blue 1, the red 4 with the blue 2, and so on, which is exactly the trick you have there in your book. No mystical insight needed.